Q10 / 160 · AP EAMCET 2014 · Chemistry

Which one of the following ions has same number of unpaired electrons as those present in $\mathrm{V}^{3+}$ ion?

  • A
    $\mathrm{Fe}^{3+}$
  • Correct answer
    B
    $\mathrm{Ni}^{2+}$
  • C
    $\mathrm{Mn}^{2+}$
  • D
    $\mathrm{Cr}^{3+}$

Solution

$\begin{aligned}{ }_{23} \mathrm{~V} & =[\mathrm{Ar}] 3 d^3 4 s^2 \\ \mathrm{~V}^{3+} & =[\mathrm{Ar}] 3 d^2 4 s^0 \Rightarrow 2 \text { unpaired electrons } \\ { }_{26} \mathrm{Fe} & =[\mathrm{Ar}] 3 d^6 4 s^2 \\ \mathrm{Fe}^{3+} & =[\mathrm{Ar}] 3 d^5 4 s^0 \Rightarrow 5 \text { unpaired electrons } \\ { }_{28} \mathrm{Ni} & =[\mathrm{Ar}] 3 d^8 4 s^2 \\ \mathrm{Ni}^{2+} & =[\mathrm{Ar}] 3 d^8 4 s^0 \Rightarrow 2 \text { unpaired electrons } \\ { }_{25} \mathrm{Mn} & =[\mathrm{Ar}] 3 d^5 4 s^2 \\ \mathrm{Mn}{ }^{2+} & =[\mathrm{Ar}] 3 d^5 \Rightarrow 5 \text { unpaired electrons } \\ { }_{24} \mathrm{Cr} & =[\mathrm{Ar}] 3 d^5 4 s^1 \\ \mathrm{Cr}^{3+} & =[\mathrm{Ar}] 3 d^3 4 s^1 \Rightarrow 3 \text { unpaired electrons }\end{aligned}$

Hence, the correct answer is option: 'B'.

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